Python 如何使用字典理解计算关键子字符串匹配的字典值小计

我想把下面的字典逻辑转换成字典理解逻辑，我该怎么做

# Example: You have this dictA as input 
dictA = {'chiken_biryani': 350, 'chiken_chilli': 300, 'motton_fry': 350, 'motton_biryani': 400, 'fish_prowns_fry': 250, 'fish_dry':300}

# Q: Print the total of each category, chiken, motton and fish items sub total values, using dictionary comprehension?
# Expected Output: {'chicken': 650, 'motton': 750, 'fish': 550}
dictB = dict()

for (k,v) in dictA.items():
    k = k.split('_')[0]
    if dictB.get(k) is None:
        dictB[k] = v
    else:
        dictB[k] = dictB.get(k)+v

print(dictB)

输出：

{'chiken': 650, 'motton': 750, 'fish': 550}

---

如果您确定订单，可以使用

itertools

中的

groupby

：

{k: sum(x[1] for x in g) for k, g in groupby(dictA.items(), lambda x: x[0].split('_')[0])}

dictA = {'chiken_biryani': 350, 'chiken_chilli': 300, 'motton_fry': 350, 'motton_biryani': 400, 'fish_prowns_fry': 250, 'fish_dry':300}

out = {k: sum(vv for kk, vv in dictA.items() if kk.startswith(k)) for k in set(k.split('_')[0] for k in dictA)}
print(out)

示例：

from itertools import groupby

# Example: You have this dictA as input 
dictA = {'chiken_biryani': 350, 'chiken_chilli': 300, 'motton_fry': 350, 'motton_biryani': 400, 'fish_prowns_fry': 250, 'fish_dry':300}

dictB = {k: sum(x[1] for x in g) for k, g in groupby(dictA.items(), lambda x: x[0].split('_')[0])}

print(dictB)
# {'chiken': 650, 'motton': 750, 'fish': 550}

---

另一种解决方案，不带

itertools

：

{k: sum(x[1] for x in g) for k, g in groupby(dictA.items(), lambda x: x[0].split('_')[0])}

dictA = {'chiken_biryani': 350, 'chiken_chilli': 300, 'motton_fry': 350, 'motton_biryani': 400, 'fish_prowns_fry': 250, 'fish_dry':300}

out = {k: sum(vv for kk, vv in dictA.items() if kk.startswith(k)) for k in set(k.split('_')[0] for k in dictA)}
print(out)

印刷品：

{'chiken': 650, 'motton': 750, 'fish': 550}

---

您可以创建一组唯一键，然后在找到该键时对字典进行迭代，求和值：

dictA = {'chiken_biryani': 350, 'chiken_chilli': 300, 'motton_fry': 350, 'motton_biryani': 400, 'fish_prowns_fry': 250, 'fish_dry':300}

key_list = set([item.split('_')[0] for item in dictA.keys()])

dictB = {unique_key: sum(el for key, el in dictA.items() if key.split('_')[0]==unique_key) for unique_key in key_list}

结果:

{'chiken': 650, 'motton': 750, 'fish': 550}

---

为什么要使用dict comp？这是可行的，但会很难看

我会使用

defaultdict

from collections import defaultdict

dict_b = defaultdict(int)

for k, v in dict_a.items():
    dict_b[k.split('_')[0]] += v